Brownian motion and Beta distribution

I am interested in the distribution of the time that the standard Brownian $W_t$ motion on $[0,1]$ satisfies the following inequality: $$W_t \ge stW(1)$$ For different values of $s$. I conjecture that the distribution is always a Beta distribution with both parameters equal (if they are Beta distributed they have to be equal because by symmetry the expected value of this time should be equal to $\frac{1}{2}$.
There are two special cases in which I can tell that the above is true: if $s=0$ then we have the usual question about the distribution of time BM spends above the $x$-axis in which case the answer is the $B(\frac{1}{2},\frac{1}{2})$ distribution. If $s=1$ the inequality can be transformed into: $$B_t=W_t-tW(1) \ge 0$$ Which asks about the distribution of the time that the Brownian bridge spends above the $x$-axis in which case the answer is $B(1,1)$ - the uniform distribution.
From the simulation I have conducted it seems that the result is true in general with $1$ being the highest parameter value. However, I don't have any proof nor any clue how to proceed.

Perhaps it may help to connect this process to a time-changed Wiener process so that you can more easily use the Wiener arcsine laws. Let

$$X_t = W_t - stW_1.$$

This is a Gaussian process with covariance function

$$c\left(u,t\right) = \min\left(u,t\right) - ut s\left(2-s\right).$$

Then the process

$$B_t = \left(1+ts\left(2-s\right)\right)X_{\frac{t}{1+ts\left(2-s\right)}}$$

is a Wiener process. If you know how much time $B$ spends above zero in various time intervals, can that help you say something about $X$?